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人気のある微分積分 >

integral of (cos(t))/(sin^4(t)-16)

解

\int \frac{cos ( t )}{sin ^{4} ( t ) - 16} d t

解

- \frac{1}{16} arctan (\frac{sin ( t )}{2}) - \frac{1}{32} ln ∣ sin (t) + 2 ∣ + \frac{1}{32} ln ∣ sin (t) - 2 ∣ + C

解答ステップ

\int \frac{cos ( t )}{sin ^{4} ( t ) - 16} d t

u置換積分法を適用する

= \int \frac{1}{u ^{4} - 16} d u

以下の部分分数を得る:

\frac{1}{u ^{4} - 16} : - \frac{1}{8 ( u ^{2} + 4 )} - \frac{1}{32 ( u + 2 )} + \frac{1}{32 ( u - 2 )}

= \int - \frac{1}{8 ( u ^{2} + 4 )} - \frac{1}{32 ( u + 2 )} + \frac{1}{32 ( u - 2 )} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{8 ( u ^{2} + 4 )} d u - \int \frac{1}{32 ( u + 2 )} d u + \int \frac{1}{32 ( u - 2 )} d u

\int \frac{1}{8 ( u ^{2} + 4 )} d u = \frac{1}{16} arctan (\frac{u}{2})

\int \frac{1}{32 ( u + 2 )} d u = \frac{1}{32} ln ∣ u + 2 ∣

\int \frac{1}{32 ( u - 2 )} d u = \frac{1}{32} ln ∣ u - 2 ∣

= - \frac{1}{16} arctan (\frac{u}{2}) - \frac{1}{32} ln ∣ u + 2 ∣ + \frac{1}{32} ln ∣ u - 2 ∣

代用を戻す

u = sin (t)

= - \frac{1}{16} arctan (\frac{sin ( t )}{2}) - \frac{1}{32} ln ∣ sin (t) + 2 ∣ + \frac{1}{32} ln ∣ sin (t) - 2 ∣

定数を解答に追加する

= - \frac{1}{16} arctan (\frac{sin ( t )}{2}) - \frac{1}{32} ln ∣ sin (t) + 2 ∣ + \frac{1}{32} ln ∣ sin (t) - 2 ∣ + C

グラフ

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